| x/(1-x2) + x2/(1-x4) +
x4/(1-x8) +........ = [1/(1-x) -1/(1-x2)
] + [1/(1-x2) -1/(1-x4) ] + [1/(1-x4) -1/(1-x8) ]
+...... Hence n th term = [1/(1-x2^n-1) -1/(1-x2^n) ]
& Sum of n terms is 1/(1-x) - 1/(1-x2^n) . If |x| < 1,
the sum up to infinity is 1/(1-x) -1 = x/ (1-x) |